Question

an engineer uses the equation below to model the height of a cannonball shot straight up from ground level. in the model the height ( y ) (in feet) is a function of ( x ), the number of seconds after the cannonball is shot. y = -16x^2 + 32x complete the parts below. (a) graph the parabola ( y = -16x^2 + 32x ). to do so, plot five points on the parabola: the vertex, two points to the left of the vertex, and two points to the right of the vertex. then click on the graph - a - function button.

Explanation:

Step1: Find the vertex of the parabola

For a quadratic function \( y = ax^2 + bx + c \), the x - coordinate of the vertex is given by \( x=-\frac{b}{2a} \). In the function \( y=-16x^{2}+32x \), \( a = - 16 \) and \( b = 32 \).
So, \( x=-\frac{32}{2\times(-16)}=-\frac{32}{-32} = 1 \).
To find the y - coordinate of the vertex, substitute \( x = 1 \) into the function: \( y=-16\times(1)^{2}+32\times(1)=-16 + 32=16 \). So the vertex is \( (1,16) \).

Step2: Find points to the left of the vertex (x = 0 and x=- 1)

• When \( x = 0 \): \( y=-16\times(0)^{2}+32\times(0)=0 \). So the point is \( (0,0) \).
• When \( x=-1 \): \( y=-16\times(-1)^{2}+32\times(-1)=-16-32=-48 \). So the point is \( (-1, - 48) \).

Step3: Find points to the right of the vertex (x = 2 and x = 3)

• When \( x = 2 \): \( y=-16\times(2)^{2}+32\times(2)=-16\times4 + 64=-64 + 64 = 0 \). So the point is \( (2,0) \).
• When \( x = 3 \): \( y=-16\times(3)^{2}+32\times(3)=-16\times9+96=-144 + 96=-48 \). So the point is \( (3,-48) \).

The five points are: vertex \( (1,16) \), left - side points \( (0,0) \), \( (-1,-48) \), right - side points \( (2,0) \), \( (3,-48) \). To graph the parabola, plot these five points \((-1,-48)\), \((0,0)\), \((1,16)\), \((2,0)\), \((3,-48)\) and then draw a smooth parabola through them.

(Note: Since the question asks to graph the parabola by plotting these points, the key is to identify these points as above. If we were to just state the points for plotting, the five points are \((-1, - 48)\), \((0,0)\), \((1,16)\), \((2,0)\), \((3,-48)\))

Answer:

The five points to plot are \((-1, - 48)\), \((0,0)\), \((1,16)\), \((2,0)\), \((3,-48)\) (and then use the graph - a - function button to draw the parabola through these points).